Calculating Tension on 500kg Beam Supported by Cable

[yankees26an]

Homework Statement

A 500 kg beam is free to pivot about its
lower end and is supported by a horizontal
cable at its upper end, as shown in the
figure. (figure has angle 35 degrees between beam and the wall)
Assuming the mass of the beam is
uniformly distributed, what is the tension in
the cable?

Homework Equations

Torque = r * F

The Attempt at a Solution

Using pivot point at the end of the beam, the torque is 0.

weight acting down on the beam is mg*l/2*sin(35). L is not givin not sure where to go from there.

 

[PhanthomJay]

Homework Statement

A 500 kg beam is free to pivot about its
lower end and is supported by a horizontal
cable at its upper end, as shown in the
figure. (figure has angle 35 degrees between beam and the wall)
Assuming the mass of the beam is
uniformly distributed, what is the tension in
the cable?

Homework Equations

Torque = r * F

The Attempt at a Solution

Using pivot point at the end of the beam, the sum of the torques from all forces is 0.

torque from weight acting down on the beam is mg*l/2*sin(35), clockwise. L is not givin not sure where to go from there.

What's the torque from the cord tension about the pivot?

 

[yankees26an]

What's the torque from the cord tension about the pivot?

you mean from the top of the beam to the bottom? :uhh:

 

[PhanthomJay]

you mean from the top of the beam to the bottom? :uhh:

yes.

 

[yankees26an]

yes.

Not sure how to find it.

 

[PhanthomJay]

Not sure how to find it.

You correctly identified the torque about the pivot from the weight force as mgL/2sintheta, that is Torque = r X F = rFsintheta, where F in this case is mg, and r is L/2, and theta is 35 degrees(the angle between the force and position vector). Now do the same for the torque about the pivot for the unknown horizontal tension force (call it T) at the top of the beam, that is, Torque from T = r X F = rFsintheta, where r is L, T is itself, and theta , the angle between T and L , is ?.. Solve for T by summing these 2 torques and setting that sum equal to 0. The 'L" term should cancel out.

 

[yankees26an]

mg*L/2*sin(35) = T*L*sin(55)

mg/2*sin(35) = T*sin(55)

(mg/2*sin(35))/sin(55) = T

T = 1.72 kN

Thanks :)